- Apple Computer introduced the iMac computer in August 1998 and Apple sold as many iMacs as they could make. One issue of great concern to Apple was whether they were expanding the base of Macintosh users or if the computers were being purchased by existing Apple customers. To investigate, ComputerWare surveyed a random sample of 500 of its iMac purchasers and found that 143 had never owned an Apple product. Construct a 95% confidence interval for the proportion of iMac purchasers who were new to Apple.
- P: What is the parameter of interest?
- \(p=\) The true proportion of iMac purchasers who were new to Apple.
- A: Show that the Assumptions and Conditions for the confidence interval have been met. (Hint: RITaLeN)
- r: We have a simple random sample of 500 of its iMac purchasers
- i: Each purchasers user status (new or old) is independent of the others
- t: 500 is less than 10% of all people who purchased iMacs.
- aLen: \(n\hat{p}=x=143 \ge 5\) and \(n(1-\hat{p})=n-x=357 \ge 5\)
- N: Have the conditions been met? What is the name of the confidence interval you will be constructing?
- The conditions have been met to construct a 95% \(z\)- interval for \(p\).
- I: Interval mechanics. Please write the formula, as symbols and with the numbers filled in. Also, give your final interval.
\[\hat{p} \pm 1.96 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=0.286\pm1.96 \sqrt{\frac{0.286(1-0.286)}{500}}=(0.246, 0.326)\]
Graphing calculator commands: stat > tests > 1-propzint
- C: Conclusion in context: Write what the interval means in the context of this problem.
- We are 95% confident that about 25 - 33% of the iMac purchasers were new to apple.
- After one year and the sale of 2,000,000 iMacs, Apple reports that one third of the new buyers of iMacs are first-time computer buyers. Does your confidence interval support their claim? Explain.
- No, 1/3 is higher than our entire confidence interval so it is not a plausible value.
- Mikki Hebl and Jingping Xu researched physician’s treatment of overweight patients. Physicians – people who are trained to treat all their patients warmly and have access to literature suggesting uncontrollable and hereditary aspects of obesity – often mirror society’s stereotypical beliefs that obese individuals are undisciplined and suffer from controllability issues. In a random sample of 38 overweight people, the doctors spent an average of 24.7 minutes with the patient. The standard deviation was 9.7 minutes. Construct a 95% confidence interval for the mean amount of time doctors spend with their overweight patients.
Mikki Hebl and Jingping Xu researched physician’s treatment of overweight patients. Physicians – people who are trained to treat all their patients warmly and have access to literature suggesting uncontrollable and hereditary aspects of obesity – often mirror society’s stereotypical beliefs that obese individuals are undisciplined and suffer from controllability issues.
In a random sample of 38 overweight people, the doctors spent an average of 24.7 minutes with the patient. The standard deviation was 9.7 minutes. Construct a 95% confidence interval for the mean amount of time doctors spend with their overweight patients.
- P: What is the parameter of interest?
- \(\mu\) = the true mean amount of time doctors spend with their overweight patients.
- A: Show that the Assumptions and Conditions for the confidence interval have been met. (Hint: RITaLeN)
- N: Have the conditions been met? What is the name of the confidence interval you will be constructing?
- The conditions have been met to construct a 95% \(t\)-interval for \(\mu\). \(df=38-1=37\)
- I: Interval mechanics. Please write the formula, as symbols and with the numbers filled in. Also, give your final interval.
- scientific calculator? - use \(df=37 \rightarrow 30\)
\[ \bar{x} \pm t^\ast \frac{s_x}{\sqrt{n}} = 24.7\pm 2.042 \cdot \frac{9.7}{\sqrt{38}}=(21.5, 27.9)\]
- C: Conclusion in context: Write what the interval means in the context of this problem.
- We are 95% confident that doctors spend on average 21.5 to 27.9 minutes with their overweight patients.
- In the same study, the mean time that doctors spent with average weight patients was 31.4 minutes. Does your confidence interval provide evidence that doctors spend less time with their overweight patients, on average?
- Yes, the entire confidence interval is less than 31.4 indicating that the true average time for overweight patients is less than for average weight patients.
- Assume that the mean time that doctors spend with their patients is normally distributed. Using the data from the study above, find the following probabilities:
- What is the probability that a doctor will spend at least 32 minutes with any one given overweight patient.
- Assume \(\mu=24.7\) and \(\sigma=9.7\).
\[P(x\ge32)=P(z\ge\dfrac{32-24.7}{9.7}\approx P(z>0.7526)\approx 0.2259\]
- Draw a picture for more points.
- In a sample of 38 overweight patients, what is the probability that the average will be over 32 minutes?
- Assume \(\mu=24.7\) and \(\sigma=9.7\).
\[P(x\ge32)=P(z\ge\dfrac{32-24.7}{\frac{9.7}{\sqrt{38}}}\approx P(z>4.64)\approx 1.75 \times 10^{-6} \approx 0.000002\]
- Draw a picture for more points.
- For each conjecture, state the null and alternative hypotheses. (Symbols only OK).
- More than 25% of adults males wax their chest hair.
- \(H_0\): \(p=.25\)
- \(H_a\): \(p>.25\)
- The average pulse rate (in beats per minutes) of females is 60.
- \(H_0\): \(\mu=60\) bpm
- \(H_a\): \(\mu\ne60\) bpm
- The mean weight of women who won Miss America titles is less than 125 lbs.
- \(H_0\): \(\mu=125\) lbs
- \(H_a\): \(\mu<125\) lbs
- Plain M&M candies have a mean weight that is at least 0.8535 g.
- \(H_0\): \(\mu=0.8535\) g
- \(H_a\): \(\mu>0.8535\) g
- The percentage of community college students receiving a pell grant is less than 20 %.
- \(H_0\): \(p=.2\)
- \(H_a\): \(p<.2\)
- Suppose that a manufacturer is testing one of its machines to make sure that the machine is producing more than 97% good parts, using a .05 significance level.
- State the null and alternative hypothesis in both words and symbols.
- \(H_0\): The machine is not producing more than 97% good parts. (\(p=0.97\))
- \(H_a\): The machine is producing more than 97% good parts. (\(p>0.97\))
- Their sample yields a p-value of 0.122. Write out the final conclusion they should make.
M: Make a decision
- M: if \(p < 0.05\) we reject \(H_0\) and if \(p > 0.05\) we fail to reject \(H_0\).
- \(p=0.122>\alpha=0.05\), so we fail to reject \(H_0\).
S: State your conclusion in context.
S: If you reject the null, state that there is strong evidence for “the alternative”. If you fail to reject the null, state there is not strong evidence for “the alternative”.
There is not strong evidence that the machine is producing more than 97% good parts.
- To determine if oxygen causes people to think more clearly, students volunteered to inhale supplemental oxygen for 10 minutes before taking a test. In fact, some received oxygen, but others (randomly assigned) were given just normal air. Test results showed that 42 of 66 students who breathed oxygen scored well, compared to only 35 of 63 students who did not get the oxygen.
- Which procedure should we use to see if there is evidence that breathing extra oxygen can help test-takers think more clearly?
- 1-proportion z-test B) 2-proportion z-test C) 1-sample t-test D) 2-sample t-test
- Do you think the conditions have been met for that test. Explain why or why not. Hint: RITaLeN
- We have random assignment to treatment and the success/failure condition is met for both groups so I would say yes.
- Many studies have been conducted to test the effects of marijuana use on mental abilities. In one such study, groups of light and heavy users of marijuana in college were tested for memory recall. 65 subjects were heavy marijuana users, with a mean memory score of 51.3 and a standard deviation of 4.5. 64 subjects were light marijuana users with a mean of 53.3 and a standard deviation of 3.6. Use a 0.05 significance level to test the claim that heavy marijuana users have a lower mean than the light users.
- P: What is the parameter of interest?
- \(\mu_H - \mu_L\) = the true difference in mean memory score for heavy and light users of marijuana.
- H: State the null and alternative hypotheses.
\(H_0:\) There is no difference in mean memory score for heavy and light users of marijuana. (\(\mu_H - \mu_L =0\))
\(H_a:\) Heavy marijuana users have a lower mean memory score than the light users. (\(\mu_H - \mu_L < 0\))
- A: Show that the Assumptions and Conditions for the test have been met. (Hint: RITaLeN)
- We don’t know if they are independent, random samples, we hope that they are at least representative of their respective populations.
- 65 & 64 are less than 10% of the population of heavy and light marijuana users respectively.
- Each sample is more than 30 so the central limit theorem applies.
- N: Have the conditions been met? What is the name of the test you will be performing? Include df.
- Conditions have been met to conduct a 2-sample \(t\) test for \(\mu_H - \mu_L\). \(df=121.9\)
- T: Test statistic. Please show your calculation and your answer.
- O: Obtain a p-value. Draw a curve and show the p-value calculations.
- M: Make a decision and S: State your conclusion in context.
- M: Since \(p=0.003 < \alpha=0.05\) we reject the null. We have strong evidence that Heavy marijuana users have a lower mean memory score than the light users.